How long until it happens: 1 ÷ p
If something happens with probability p on each try, you expect to wait 1/p tries. The wait does not shrink while you wait, several successes take several waits, and collecting every type takes far longer than the count suggests.
- Compute the expected wait for a first success and the chance you are still waiting after k tries
- Explain memorylessness: why a long silence does not make the event 'due'
- Add waits for the m-th success
- Sum the coupon-collector stages and see why the last one dominates
The intuition
A signal fires on any given day with probability 10%. How long until it fires? Ten days on average, 1 ÷ p, and that number has a property people find hard to accept: after ten quiet days the expected remaining wait is still ten days. Independent trials carry no memory. 'It is overdue' is a statement about you, not about the process.
The idea chains. The wait for the third firing is three waits of ten days, because expectations add. And the wait to see every one of N regime types is a sum of waits whose success probabilities fall as the collection fills: the first new type arrives on day one, the last needs a 1-in-N shot every day. With six types the whole collection takes about 14.7 days, and the last type alone takes six of them.
E[trials to first success] = 1 ÷ p. P(still waiting after k) = (1 − p)ᵏ; P(first success on trial k) = (1 − p)ᵏ⁻¹ p. E[trials to the m-th success] = m ÷ p. Coupon collector with N types: N × (1 + 1/2 + … + 1/N). From P(silent for k) = x: p = 1 − x^(1/k).
Why it works
- The conventions here: independent trials with a fixed success probability p. The count of trials up to and including the first success is geometric; up to the m-th success, negative binomial. All expectations are exact.
- Why 1 ÷ p. Over many trials a share p of them are successes, so successes are 1/p trials apart on average. The distribution has a long right tail: the median wait is shorter than the mean, because a few very long waits pull the mean up.
- Memorylessness. P(wait > k + j | wait > k) = P(wait > j). The past failures are already spent; the future looks exactly as it did on day one.
- Several successes. Each success is a fresh geometric wait and expectations add, so m ÷ p. Independence is not even needed for the addition.
- Coupon collector. With j types already seen, the next new one arrives with probability (N − j) ÷ N, so that stage takes N ÷ (N − j) days. Summing from j = 0 to N − 1 gives N × H_N, and the final stage, N days, is the biggest single term.
- Run it backwards. A silence of k days has probability (1 − p)ᵏ. Given that probability, the k-th root recovers the daily miss rate, and one minus it is p.
| Expected wait: 1 ÷ 10% | 10 days |
| P(no fire in 5 days): 0.9⁵ | 59.05% |
| P(first fire on day 5): 0.9⁴ × 0.1 | 6.56% |
| Expected days to 3 firings: 3 ÷ 10% | 30 days |
| Coupon collector, 6 regimes: 6 × (1 + 1/2 + … + 1/6) | 14.7 days |
| Of which the last regime alone: 6 ÷ 1 | 6 days |
After 5 silent days the expected further wait is still 10 days. The median wait is 7 days, the first k with 0.9ᵏ ≤ ½, below the mean of 10.
The formulas
The geometric mean.
k failures; or k − 1 failures then a success.
m waits of 1 ÷ p; expectations add.
One stage per new type, each a geometric wait.
The daily rate from the chance of a silence.
Worked example
One over the daily probability. The follow-up asks whether the median wait is shorter or longer than the mean.
See it move
Same signal. Change how often it fires, how many days you have waited, how many firings you need, and how many regime types there are to collect.
- Raise the daily probability. The expected wait falls, the chance of still waiting falls, and the wait for several firings falls with it.
- Wait more days. The chance of still waiting falls, and so does the chance of the first firing landing exactly on that day; the expected further wait, not shown because it never moves, stays at 1 ÷ p.
- Add regime types. The days to collect them all rise, and so do the days the last type alone takes, one for each type.
Run it backwards
The chance of a silence of a given length is known. What is the daily probability, and the expected wait?
(1 − p)ᵏ is the silence probability. Take the k-th root for the daily miss rate; p is one minus that; the expected wait is 1 ÷ p.
It is the same inverse as the coin's bias in the last chapter: a run of misses, solved for the rate.
Traps
Say it in the interview
“A signal fires on 10% of days and has been quiet for two weeks. When will it fire?”
Check yourself
5 fresh questions, with new numbers. Answer each one correctly to finish the chapter. Get one wrong and you will see the full working, then you can try it again with new numbers.
Answers within 1% are marked right. Type the number; $, %, x and M are fine. First tries count toward Learned: the topic is Learned once every chapter is done and 75% of first tries were right.
- Expected wait = 1 ÷ p, and it does not shrink while you wait.
- Still waiting after k: (1 − p)ᵏ. First success on k: (1 − p)ᵏ⁻¹p.
- m successes take m ÷ p; every type takes N × H_N, and the last type takes N.
- A silence of k days has probability (1 − p)ᵏ; take the k-th root to get the rate back.